Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000000000")
using namespace std;
int n;
int T[100001], P[100001][20], L[100001];
vector<int> adj[100001], citizen[100001];
bool vis[100001];
vector<int> PC[100001][20];
void DFS(int cur) {
  vis[cur] = 1;
  for (int i = 0; i < adj[cur].size(); i++) {
    int nx = adj[cur][i];
    if (vis[nx]) continue;
    T[nx] = cur;
    L[nx] = L[cur] + 1;
    DFS(nx);
  }
}
vector<int> Merge(vector<int> a, vector<int> b) {
  vector<int> ret;
  int i = 0, j = 0;
  while (i < a.size() && j < b.size() && ret.size() < 10) {
    if (a[i] < b[j])
      ret.push_back(a[i++]);
    else
      ret.push_back(b[j++]);
  }
  while (i < a.size() && ret.size() < 10) {
    ret.push_back(a[i++]);
  }
  while (j < b.size() && ret.size() < 10) {
    ret.push_back(b[j++]);
  }
  return ret;
}
void process3() {
  int i, j, N = n;
  for (i = 0; i < N; i++)
    for (j = 0; 1 << j < N; j++) P[i][j] = -1;
  for (i = 0; i < N; i++) {
    P[i][0] = T[i];
    for (int j = 0; j < citizen[i].size(); j++)
      PC[i][0].push_back(citizen[i][j]);
    sort((PC[i][0]).begin(), (PC[i][0]).end());
    while (PC[i][0].size() > 10) PC[i][0].pop_back();
  }
  for (j = 1; 1 << j <= N; j++)
    for (i = 0; i < N; i++)
      if (P[i][j - 1] != -1) {
        P[i][j] = P[P[i][j - 1]][j - 1];
        PC[i][j] = Merge(PC[i][j - 1], PC[P[i][j - 1]][j - 1]);
      }
}
int query(int p, int q) {
  int tmp, log, i, N = n;
  if (L[p] < L[q]) tmp = p, p = q, q = tmp;
  for (log = 1; 1 << log <= L[p]; log++)
    ;
  log--;
  for (i = log; i >= 0; i--)
    if (L[p] - (1 << i) >= L[q]) {
      p = P[p][i];
    }
  if (p == q) return p;
  int o = 0;
  for (i = log; i >= 0; i--)
    if (P[p][i] != -1 && P[p][i] != P[q][i]) p = P[p][i], q = P[q][i], o = i;
  return T[p];
}
vector<int> GetPath(int u, int dis) {
  vector<int> ret;
  for (int i = 0; i < 20; i++)
    if (dis & (1 << i)) {
      ret = Merge(ret, PC[u][i]);
      u = P[u][i];
    }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  int u, v, m, q, a;
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 0; i < m; i++) {
    cin >> u;
    u--;
    citizen[u].push_back(i + 1);
  }
  T[0] = -1;
  L[0] = 1;
  DFS(0);
  process3();
  while (q--) {
    cin >> u >> v >> a;
    u--, v--;
    int lca = query(u, v);
    vector<int> A = GetPath(u, L[u] - L[lca]);
    vector<int> B = GetPath(v, L[v] - L[lca] + 1);
    A = Merge(A, B);
    cout << min(int(A.size()), a) << ' ';
    for (int i = 0; i < min(a, (int)A.size()); i++) cout << A[i] << ' ';
    cout << endl;
  }
  return 0;
}